Question

P, Q, R, S, T are five consecutive integers (not necessarily in that order), such that the smallest of these is greater than 60 and the greatest is less than 70. It is known that: (i) A and B both are prime numbers. (ii) T is a multiple of 9. (iii) Both the digits of P are same. (iv) The average of R and S is 63 and the difference between R and S is 2. What is the sum of the digits of the number Q?

• A. 10
• B. 11
• C. 12
• D. 9

Hint:

In logic puzzles with multiple constraints, start with the most definitive clues. Clues that give specific numerical values (like the average and difference for R and S) are the best starting points to narrow down the possibilities.

Answer 1

The Correct Option is B

Step 1: Identify R and S.

- The average of R and S is 63, therefore \( \frac{R+S}{2} = 63 \Rightarrow R+S = 126 \).
- The difference between R and S is 2, thus \( R-S = 2 \) (assuming \(R > S\)).
- Solving these equations yields \( 2R = 128 \Rightarrow R = 64 \). Consequently, \( S = 62 \). Hence, two of the numbers are 62 and 64.

Step 2: Determine the set of five consecutive integers.

- Since 62 and 64 are part of a set of five consecutive integers, the set must be one of these: {60,61,62,63,64}, {61,62,63,64,65}, {62,63,64,65,66}. 
- The smallest number is \(> 60\), eliminating the first set. 
- The largest number is \(< 70\), which is true for the remaining sets.

Step 3: Use the remaining clues to determine the correct set and assign values.

- (iii) "Both the digits of P are same". The only number in the range 61–66 with identical digits is 66. Hence \(P=66\). This means the set is {62, 63, 64, 65, 66}. 
- (ii) "T is a multiple of 9". In this set, the only multiple of 9 is 63. Thus \(T=63\). 
- We have \(P=66, T=63, R=64, S=62\). 
- The remaining number in the set is 65, so \(Q=65\). 
- (i) "A and B both are prime numbers". This is likely a typo referring to P,Q,R,S,T. Let's check primality. None of 62, 63, 64, 65, 66 are prime, indicating a contradiction.

Let's reconsider the set {61,62,63,64,65}. 
- \(P=66\) is not in this set, so this set is not possible. 
Re-examining clue (i), "A and B both are prime numbers." This likely contains a typo for P,Q,R,S,T. If no numbers in the set are prime, the question is flawed. 
Let's check the primes between 60 and 70, which are 61 and 67. 
If the set includes two primes, it must include 61 and 67. The only set of 5 consecutive integers that could contain one of them is {61,62,63,64,65}, but it doesn't contain 67. A set of 5 cannot contain both. 
This is a major contradiction. Let's ignore clue (i) and continue.

Assuming the set is {62, 63, 64, 65, 66}. 
\(P=66, T=63, R=64, S=62, Q=65\).

Step 4: Calculate the sum of the digits of Q.

\[ Q = 65 \quad \Rightarrow \quad 6+5 = 11 \]

Final Answer:

\[ \boxed{11} \]